What makes the matter to exist in any one of three state of matter? Why should the properties of solids, liquids and gases be different? Before finding a solution for the above mentioned questions, let us recollect that: All matter is made up of very small particles. These particles may be atoms, ions or molecules. […]
What factors governs the state of matter? Read More »
We are given with, Molarity = 0.001 M Volume = 100 ml = 0.1 L Number of moles of H2SO4 = Molarity*Volume = 0.001 M*0.1 L = 0.0001 mole Dissociation of sulphuric acid occurs as: H2SO4 ⟶ 2H+ + SO42- From the above equation, 1 molecule H2SO4 given 1 SO42- ion then,
Calculate the number of sulphate ions in 100 ml of 0.001 M H2SO4 solution. Read More »
Given mass of CO2 = 200 mg = 0.2 g Molar mass of CO2 = 44 g/mol Number of moles of 200 mg CO2 = Given mass/Molar mass of CO2 = 0.2 g/44 g mol-1 = 0.00454 mole Number of molecules of CO2 removed = 1021 Number of moles of CO2 removed = 1021/(6.022*1023) =
From 200 mg of CO2, 10^21 molecules are removed. How many moles of CO2 are left? Read More »
Q. Assertion : The element having with electronic configuration 1s2 2s2 2p6 3s2 and 1s2 2s2 belong to the same group Reason: Both have same number of electrons in the valence shell Solution: Elements having same number of valence shell are present in same group. Here, each elements have 2 valence electrons.
Assertion-Reason based electronic configuration Read More »
30% w/v H2SO4 means 30 g H2SO4 in 100 ml solution. Mass of H2SO4 = 30 g Density of H2SO4 = 1.05 g/ml Volume of H2SO4 = Mass/Density = (30 g)/(1.05g/ml) = 28.57 ml Volume of solution = 100 ml Density of solution = 1.5 g/ml Mass of solution = Volume*Density = (100 ml)*(1.5 g/ml) = 150
In KMnO4, oxidation state of Mn is +7. As the oxidation state of atom increases, the element become more electronegative in nature. Mn gain 5 electron in acidic medium and reduces to +2 oxidation state. Gain of 5 electron shows that the electronegativity of Mn has increased. Therefore, KMnO4 is a good oxidizing agent. Note:
Why KMnO4 act as a good oxidizing agent in acidic medium? Read More »
Question: If E° for copper electrode is +0.34 V, how will you calculate its emf value when the solution in contact with it is 0.1 M in copper ions? How does emf for copper electrode change when concentration of Cu2+ ions in the solution is decreased? Solution: The emf of an electrode when dipped in
E° for copper electrode is +0.34 V, how will you calculate its emf value? Read More »
Molar conductivity increases with increase in volume and when volume increases the concentration of solution decreases (because no.of ions in the solution decrease) which makes the solution dilute. On dilution when concentration approaches zero, the molar conductivity is known as limiting molar conductivity. In case of strong electrolytes the ionisation is easy and
How can you determine limiting molar conductivity for strong and weak electrolyte? Read More »
Solution: Given, [A]=0.1 mol L-1, [B]=0.2 mol L-1, k = 2.0×10-6 mol-2 L2 s-1Rate =k[A][B]2 = (2.0x 10-6 ) x (0.1 mol L-1) x (0.2 mol L-1) 2 = 8.0 x 10-9 mol-2 L2 s-1. The concentration of A is reduced to (0.1 – 0.06) = 0.04 mol L-1.Now according to the equation 1 mole of A reacts with 2 mole of B. So, 0.04
Calculate the initial rate of reaction. Read More »
Let x g of each gas is mixed. Molar mass of ethane = 30 g/mol Molar mass of hydrogen = 2 g/mol Mole = Given mass/Molar mass Moles of ethane = (x/30) mol Moles of hydrogen = (x/2) mol So, the fraction of total pressure exerted by hydrogen
